Quick questions on Solubility product Ksp explained: HSC Chemistry Module 5

For a sparingly soluble ionic compound, dissolution is an equilibrium between the solid and its ions in solution:

For any solution (not necessarily at equilibrium), the same expression evaluated using current concentrations is the ionic product Q.

Adding a common ion to a saturated solution decreases solubility. For example, adding NaCl to a saturated solution of AgCl increases $[Cl^-]$, so $[Ag^+]$ must decrease to keep $Q = K_{sp}$. Some AgCl precipitates.

Note that $PbCl_2$ has a larger Ksp than AgCl but the relationship to solubility is not direct (different stoichiometries). For $AB$ salts, $s = \sqrt{K_{sp}}$. For $AB_2$ salts, $s = \sqrt[3]{K_{sp}/4}$.

$NaCl$ fully dissociates, so $[Cl^-]$ from the salt is 0.10 mol/L before any AgCl dissolves. Let $s'$ be the molar solubility of AgCl in this solution.

Because Ksp is tiny and $[Cl^-]$ from NaCl is much larger than $s'$, $0.10 + s' \approx 0.10$.

In pure water $s = 1.34 \times 10^{-5}$ mol/L; in 0.10 mol/L NaCl, $s' = 1.8 \times 10^{-9}$ mol/L. The solubility falls by a factor of ~7400. Adding the common ion drives the equilibrium to the left (toward solid), exactly as Le Chatelier's principle predicts.

$s' / 0.10 = 1.8 \times 10^{-8}$, far below 5%. Valid.

For $PbCl_2$, $[Cl^-]$ is squared. Lose this and the calculation collapses.

$[Cl^-] = 2s$, not $s$.

The salt with the larger Ksp is not necessarily more soluble. Convert to molar solubility before comparing.

When two solutions are combined, the concentrations of each ion are diluted by the dilution factor (volume mixed / total volume). Most "will a precipitate form" questions test this.

Q changes as the system approaches equilibrium. Only Ksp is constant (at a given temperature).